Solve zero of a trigonometric equation This is the equation I have:
$$\sin \left(\frac{\pi t+\frac{\pi }{2}}{2}\right)-\sin \left(\frac{\pi t}{2}\right)=0$$
Also the variable t has a requirement $0\le t\le \frac{3}{2}$. I have tried $\frac{\pi t+\frac{\pi }{2}}{2}=\frac{\pi t}{2}$ but when trying to solve for t, it t gets removed. I'm not sure what to do.
 A: Just use the identity
$$
\sin (a+b)=\sin(a) \cos (b) +\cos(a) \sin (b).
$$
Then, in your case, you reduce your equation to:
$$
\sin(\tfrac{\pi}{2}x) (\tfrac{\sqrt{2}-2}{2})+\tfrac{\sqrt{2}}{2}\cos (\tfrac{\pi}{2}x)=0
$$
or, equivalently
$$
\tan y = \tfrac{\sqrt{2}}{2-\sqrt{2}}
$$
where $y=\tfrac{\pi}{2}x$.
This yields that the unique solution in $(0,\tfrac{3}{2})$ is $x=\tfrac{3}{4}$.
A: Take $(\pi t/2)=x$. Use the derived formula $$\sin A-\sin B=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac {A+B}{2}\right)$$
to get
$$\sin \left(x+\frac{\pi}{4}\right)-\sin x=0
\Rightarrow2\sin\frac{\pi}{8}\cos\left(x+\frac{\pi}{8}\right)=0\\\Rightarrow\cos\left(\frac{\pi}{2}\left[ t+\frac{1}{4}\right]\right)=0$$
We know that $\cos (n\pi)=0$ (only) for all odd multiples of $\pi/2$ i.e.,
$$t+\frac{1}{4}=2n+1\Rightarrow t=2n+\frac{3}{4}\quad (n\in\mathbb{Z}).$$
This is the general solution. If the minimum positive value is desired, put $n=0$.
NOTE. If you can't recall this formula, derive it. Hint:
$$\sin(X+Y)-\sin(X-Y)=2\sin Y\cos X$$
Put $X=(A+B)/2$ and $Y=(A-B)/2$.
